CODEWITHSUNDEEP
assemblyintelx86
justbrianr
justbrianr

Reputation: 73

Simple division function in IA32 assembly

I am working on an assignment for my class. My C++ code calls the _Divide function to take in 2 values to divide and return into 'Result'. To start I move -1 into eax. Once I'm done with the process, my eax keeps returning '-1' as the value. What could I be doing wrong? Here's my assem code:

public _Divide
.386
.model flat
.code
_Divide   proc
      mov eax, -1
      mov ebx, 1

      cdq
      idiv ebx
      ret
_Divide   endp
   end

Here's my C++ code

#include <iostream>

using namespace std;

extern "C" long Divide (long, long, long *);
void main ()
{
long    Result;
long    Remainder;
long    Dividend;
long    Divisor;
 do {
  cout << "Enter Dividend" << endl; 
  cin >> Dividend;
  cout << "Enter Divisor" << endl;
  cin >> Divisor;
  Result = Divide (Dividend, Divisor, &Remainder);
  cout << "Result is " << Result << " and Remainder is " << Remainder << endl;
 } while ((Result >= 0) || (Remainder != 0));
 Result = Divide (Dividend, Divisor, 0);
 cout << "Result is " << Result << " and Remainder is not used" << endl;
}

Thanks.

Upvotes: 1

Views: 4148

Answers (2)

Carl Norum
Carl Norum

Reputation: 225272

Your code divides -1 by 1. The answer is -1, so that's what you return. Let's break down the code (I'll just put inline comments):

mov eax, -1      ; Put -1 in eax
mov ebx, 1       ; put  1 in ebx

cdq              ; sign extend eax -> edx:eax = -1
idiv ebx         ; divide edx:eax (-1) by ebx (1)
                 ;   result goes in eax, so now eax = -1
ret              ; return eax (-1) to caller

If you want to divide arguments passed to this function, you need to access them somehow. You didn't show us the C++ function signature, so I can't help you with any more details than that.

Hint: If you're writing IA32 code, which it looks like you're doing, the arguments will be on the stack.

Upvotes: 4

Andrey
Andrey

Reputation: 60115

If you want to write function in assembly you should learn about calling conventions first: http://en.wikipedia.org/wiki/X86_calling_conventions . Basically calling convention is the set of agreement between caller and function on how they will pass and return values.

Usual convention for C++ is usually __cdecl or __stdcall. Both require return value to be passed via EAX. Since EAX always have -1 this is what you get.

Upvotes: 2

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